The graph below models the velocity of a small train as it moves on a straight track for 20 seconds - AQA - A-Level Maths Mechanics - Question 14 - 2017 - Paper 2

To calculate the total distance travelled, we can use the area under the velocity-time graph. The graph has three distinct sections:

1. Acceleration phase: From t = 0 to t = 8 seconds (velocity from 0 to 8 m/s)

   The area is a triangle:

   $S_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 8 = 32 \text{ m}$

2. Constant velocity phase: From t = 8 to t = 16 seconds (velocity of 8 m/s)

   The area is a rectangle:

   $S_2 = \text{base} \times \text{height} = 8 \times 8 = 64 \text{ m}$

3. Deceleration phase: From t = 16 to t = 20 seconds (velocity from 8 m/s to 0 m/s)

   The area is a trapezium (or a triangle since the final velocity is zero):

   $S_3 = \frac{1}{2} \times \text{base} \times \text{(height}_1 + \text{height}_2) = \frac{1}{2} \times 4 \times (8 + 0) = 16 \text{ m}$

Now, adding these areas together gives:

$\text{Total distance} = S_1 + S_2 + S_3 = 32 + 64 + 16 = 112 \text{ m}$